$(1+\tan A)(1+\tan B)=2$
$1+\tan A+\tan B+\tan A\tan B=2$
$\tan A+\tan B=1-\tan A.\tan B$
$\large\frac{\tan A+\tan B}{1-\tan A.\tan B}$$=1$
$\tan(A+B)=1=\tan\large\frac{\pi}{4}$
$\Rightarrow A+B=n\pi+\large\frac{\pi}{4}$ where $n=0,\pm 1,\pm 2$
Hence (A) is the correct answer.